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@AndriyLazorenko
Created February 17, 2017 12:46
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import sys
import math
from Week4.closest.utils.timer import timer
@timer
def closest_split_pair(p_x, p_y, delta, best_pair):
ln_x = len(p_x) # store length - quicker
mx_x = p_x[ln_x // 2][0] # select midpoint on x-sorted array
# Create a subarray of points not further than delta from
# midpoint on x-sorted array
s_y = [x for x in p_y if mx_x - delta <= x[0] <= mx_x + delta]
best = delta # assign best value to delta
ln_y = len(s_y) # store length of subarray for quickness
for i in range(ln_y - 1):
for j in range(i+1, min(i + 7, ln_y)):
p, q = s_y[i], s_y[j]
dst = dist(p, q)
if dst < best:
best_pair = p, q
best = dst
return best_pair[0], best_pair[1], best
@timer
# Impossible to do this any faster
def dist(p1, p2):
return math.sqrt((p1[0] - p2[0]) ** 2 + (p1[1] - p2[1]) ** 2)
@timer
# This one seems optimized
def brute(ax):
mi = dist(ax[0], ax[1])
p1 = ax[0]
p2 = ax[1]
ln_ax = len(ax)
if ln_ax == 2:
return p1, p2, mi
for i in range(ln_ax-1):
for j in range(i + 1, ln_ax):
if i != 0 and j != 1:
d = dist(ax[i], ax[j])
if d < mi: # Update min_dist and points
mi = d
p1, p2 = ax[i], ax[j]
return p1, p2, mi
@timer
def solution(x, y):
a = list(zip(x, y)) # This produces list of tuples
ax = sorted(a, key=lambda x: x[0]) # Presorting x-wise
ay = sorted(a, key=lambda x: x[1]) # Presorting y-wise
p1, p2, mi = closest_pair(ax, ay) # Recursive D&C function
return mi
@timer
def closest_pair(ax, ay):
ln_ax = len(ax) # It's quicker to assign variable
if ln_ax <= 3:
return brute(ax) # A call to bruteforce comparison
mid = ln_ax // 2 # Division without remainder, need int
Qx = ax[:mid] # Two-part split
Rx = ax[mid:]
# Determine midpoint on x-axis
qx = set(Qx)
Qy = list()
Ry = list()
for x in ay:
if x in qx:
Qy.append(x)
else:
Ry.append(x)
# Call recursively both arrays after split
(p1, q1, mi1) = closest_pair(Qx, Qy)
(p2, q2, mi2) = closest_pair(Rx, Ry)
# Determine smaller distance between points of 2 arrays
if mi1 <= mi2:
d = mi1
mn = (p1, q1)
else:
d = mi2
mn = (p2, q2)
# Call function to account for points on the boundary
(p3, q3, mi3) = closest_split_pair(ax, ay, d, mn)
# Determine smallest distance for the array
if d <= mi3:
return mn[0], mn[1], d
else:
return p3, q3, mi3
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